Difference between revisions of "2000 SMT/Algebra Problems/Problem 2"
| Line 1: | Line 1: | ||
| − | Problem 2 | + | ==Problem 2== |
| − | Evaluate 2000^3 - 1999(2000^2) - 1999^2(2000) + 1999^3 | + | Evaluate <math>2000^3 - 1999(2000^2) - 1999^2(2000) + 1999^3</math> |
| − | Solution 1 - SUBMITTED BY HOWDOI_YT | + | ==Solution 1 - SUBMITTED BY HOWDOI_YT== |
| − | I can write the first 1999 as 2000-1, and the 3rd 2000 as 1999+1; | + | I can write the first <math>1999</math> as <math>2000-1</math>, and the 3rd <math>2000</math> as <math>1999+1</math>; |
| − | 2000^3 - 2000^2(2000-1) - 1999^2(1999+1) + 1999^3 = 2000^3 - 2000^3 + 2000^2 - 1999^3 - 1999^2 + 1999^3 = 2000^2 - 1999^2 = (2000 - 1999)(2000 + 1999) = | + | <math>2000^3 - 2000^2(2000-1) - 1999^2(1999+1) + 1999^3</math> |
| − | + | <math>= 2000^3 - 2000^3 + 2000^2 - 1999^3 - 1999^2 + 1999^3</math> | |
| + | <math>= 2000^2 - 1999^2</math> | ||
| + | <math>= (2000 - 1999)(2000 + 1999)</math> | ||
| + | <math>= 3999 \blacksquare</math> | ||
Revision as of 15:13, 18 April 2023
Problem 2
Evaluate 
Solution 1 - SUBMITTED BY HOWDOI_YT
I can write the first 
as 
, and the 3rd 
as 
;
